Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve(int test_number);
int main() {
  cout.setf(ios::fixed);
  cout.precision(9);
  cerr.setf(ios::fixed);
  cerr.precision(3);
  int n = 1;
  for (int i = 0; i < n; i++) {
    solve(i + 1);
  }
  return 0;
}
const int MAXN = 100100;
const int LOG = 18;
const int CNT = 11;
int buf[2 * CNT];
int buf1[2 * CNT];
int buf2[2 * CNT];
int buf3[4 * CNT];
int buf4[4 * CNT];
int n, m;
vector<int> g[MAXN];
vector<int> x[MAXN];
int up[MAXN][LOG];
int mn[MAXN][LOG][CNT];
int lev[MAXN];
int timer;
int tin[MAXN];
int tout[MAXN];
inline void calc_mn(int v, int i) {
  if (lev[v] < (1 << (i - 1))) {
    for (int j = 0; j < CNT; j++) mn[v][i][j] = mn[v][i - 1][j];
  } else {
    merge(mn[v][i - 1], mn[v][i - 1] + CNT, mn[up[v][i - 1]][i - 1],
          mn[up[v][i - 1]][i - 1] + CNT, buf);
    for (int j = 0; j < CNT; j++) mn[v][i][j] = buf[j];
  }
}
inline void dfs(int v, int pred = 0, int level = 0) {
  for (int i = 0; i < LOG; i++) {
    for (int j = 0; j < CNT; j++) mn[v][i][j] = ((int)(1e+9));
  }
  lev[v] = level;
  tin[v] = timer++;
  up[v][0] = pred;
  int cnt = 0;
  for (auto &p : x[v]) {
    mn[v][0][cnt] = p;
    cnt++;
    if (cnt == CNT) break;
  }
  for (int i = 1; i < LOG; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    calc_mn(v, i);
  }
  for (auto &to : g[v]) {
    if (to != pred) dfs(to, v, level + 1);
  }
  tout[v] = timer++;
}
inline bool upper(int u, int v) {
  return tin[u] <= tin[v] && tout[v] <= tout[u];
}
inline int get_lca(int v, int u) {
  if (upper(v, u)) return v;
  if (upper(u, v)) return u;
  for (int i = LOG - 1; i >= 0; i--) {
    if (!upper(up[v][i], u)) v = up[v][i];
  }
  return up[v][0];
}
inline void get_path(int v, int lca, int *buf) {
  for (int i = 0; i < 2 * CNT; i++) buf[i] = ((int)(1e+9));
  for (int i = LOG - 1; i >= 0; i--) {
    if (!upper(up[v][i], lca)) {
      merge(buf, buf + CNT, mn[v][i], mn[v][i] + CNT, buf4);
      v = up[v][i];
      for (int j = 0; j < CNT; j++) buf[j] = buf4[j];
    }
  }
  merge(buf, buf + CNT, mn[v][0], mn[v][0] + CNT, buf4);
  for (int j = 0; j < CNT; j++) buf[j] = buf4[j];
}
inline void print(int *buf, int lca, int cnt) {
  merge(buf, buf + CNT, mn[lca][0], mn[lca][0] + CNT, buf4);
  for (int i = 0; i < 2 * CNT; i++) buf[i] = buf4[i];
  int res = 0;
  while (buf[res] != ((int)(1e+9))) {
    res++;
    if (res == cnt) break;
  }
  res = min(res, cnt);
  cout << res << ' ';
  for (int i = 0; i < res; i++) cout << buf[i] << ' ';
  cout << '\n';
}
void solve(int test_number) {
  ios_base::sync_with_stdio(false);
  cout.tie(nullptr);
  int q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int pos;
    cin >> pos;
    pos--;
    x[pos].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    sort(x[i].begin(), x[i].end());
  }
  dfs(0);
  for (int i = 0; i < q; i++) {
    int u, v, cnt;
    cin >> u >> v >> cnt;
    u--, v--;
    int lca = get_lca(u, v);
    get_path(v, lca, buf1);
    get_path(u, lca, buf2);
    if (u == v) {
      cout << min(cnt, (int)x[u].size()) << ' ';
      for (int j = 0; j < min(cnt, (int)x[u].size()); j++)
        cout << x[u][j] << ' ';
      cout << '\n';
    } else if (u == lca) {
      print(buf1, lca, cnt);
    } else if (v == lca) {
      print(buf2, lca, cnt);
    } else {
      merge(buf1, buf1 + 2 * CNT, buf2, buf2 + 2 * CNT, buf3);
      print(buf3, lca, cnt);
    }
  }
}